Question

The Ka for benzoic acid is 6.5 × 10−5. Calculate the pH of a 0.12 M benzoic acid solution.

Answer 1

Answer:

pH=4.05

Explanation:

C7H6O2  -> C7H6O-  + H+

ka= [C7H6O-] [H+]/[C7H6O2]

During equilibrium

[C7H6O-]= [H+]= x^2

[C7H6O2]=0.12-x

Replace

ka= x^2/0.12-x

6.5 x10^-5= x^2/0.12-x

7.8x10^-6 - 6.5 x10^-5x=x^2

x^2+ 6.5 x10^-5x - 7.8x10^-6

Solution of quadratic equation

x=8.8 x10^-5

pH= -log [H+]= -log 8.8 x10^-5=4.05